A fifty pound child is sitting at the end of a seesaw, six feet from it's center.?

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How far from the center on the other end must a sixty pound child sit in order to balance the see saw?

2006-11-29 14:59:17 · 4 answers · asked by amolegends 1 in Science & Mathematics ➔ Engineering

50 Lbs X 6Ft= 60 LBS X ? Ft
Multiply
300 Ft LBS= 60 LBS X ? Ft
divide
300 Ftlbs/ 60 LBS = ?Ft
5Ft= ?ft

So 5 Ft

2006-11-29 15:17:39 · answer #1 · answered by shadouse 6 · 0⤊ 0⤋

50 x 6 = d x 60 sum moments about center of seesaw

d = (50 x 6)/ 60

2006-11-29 15:03:48 · answer #2 · answered by Roadkill 6 · 0⤊ 0⤋

0 distance, cause the gravity always in the center

2006-11-29 17:04:38 · answer #3 · answered by HARP C 1 · 0⤊ 0⤋

5ft

2006-11-29 15:18:02 · answer #4 · answered by julia k 2 · 0⤊ 0⤋